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Solving Systems by Substitution
Solve the following system of equations by using
substitution.
4x + 5y = 7
2x – 3y = 9
Step 1: Solve one of the equations for a variable.
Solve the 2nd equation for x.
2x – 3y = 9
2x = 3y + 9
x = 1.5y + 4.5
Step 2: Substitute this equation into the first
equation and then solve for y.
4x + 5y = 7
(Original equation)
4(1.5y + 4.5) + 5y = 7
(With substitution)
6y + 18 + 5y = 7
11y = -11
y = -1
Step 3: Substitute the y value into the equation
from step 1.
x = 1.5y + 4.5
x = 1.5(-1) + 4.5
x = -1.5 + 4.5
x=3
Solve the system.
-2x + y + 3z = 20
-3x + 2y + z = 21
3x – 2y + 3z = -9
Step 1: Select an equation and solve it for a
variable. The 1st or 2nd would be good choices
because one of their variables has a coefficient of 1.
Let’s solve the first equation for y.
-2x + y + 3z = 20
y = 2x – 3z + 20
Step 2: Substitute this equation into the other two
equations. Simplify.
-3x + 2y + z = 21
3x – 2y + 3z = -9
-3x + 2(2x – 3z + 20) + z = 21
-3x + 4x – 6z + 40 + z = 21
3x – 2(2x – 3z + 20) + 3z = -9
3x – 4x + 6z – 40 + 3z = -9
x – 5z = -19
-x + 9z = 31
Now we have a system of two equations with two
variables.
x – 5z = -19
-x + 9z = 31
Step 3: Solve one of the equations for a variable.
Choose the 1st and solve for x.
x – 5z = -19
x = 5z – 19
Step 4: Substitute this equation into the other.
Solve for the variable.
-x + 9z = 31
-(5z – 19) + 9z = 31
-5z + 19 + 9z = 31
4z = 12
z=3
Step 5: Substitute the value of z into the equation
for x.
x = 5z – 19
x = 5(3) – 19
x = 15 – 19
x = -4
Step 6: Substitute the values of x and z into the
equation for y.
y = 2(-4) – 3(3) + 20
y = -8 – 9 + 20
y=3
The solution to the system is:
x = -4,
y = 3,
z=3
As an order triple: (-4, 3, 3)
Solve the system.
x + 2y + z = 8
2x + y – z = 4
x + y + 3z = 7
Step 1: Select an equation and solve it for a
variable. Any of the 3 would be good choices
because one of their variables has a coefficient of 1.
Let’s solve the first equation for x.
x + 2y + z = 8
x = -2y – z + 8
Step 2: Substitute this equation into the other two
equations. Simplify.
2x + y – z = 4
x + y + 3z = 7
2(-2y – z + 8) + y – z = 4
(-2y – z + 8) + y + 3z = 7
-4y – 2z + 16 + y – z = 4
-2y – z + 8 + y + 3z = 7
-3y – 3z = -12
-y + 2z = -1
Now we have a system of two equations with two
variables.
-3y – 3z = -12
-y + 2z = -1
Step 3: Solve one of the equations for a variable.
Choose the 2nd and solve for y.
-y + 2z = -1
y = 2z + 1
Step 4: Substitute this equation into the other.
Solve for the variable.
-3y – 3z = -12
-3(2z + 1) – 3z = -12
-6z – 3 – 3z = -12
-9z = -9
z=1
Step 5: Substitute the value of z into the equation
for y.
y = 2z + 1
y = 2(1) + 1
y=2+1
y=3
Step 6: Substitute the values of y and z into the
equation for x.
x = -2y – z + 8
x = -2(3) – (1) + 8
x = -6 – 1 + 8
x=1
The solution to the system is:
x = 1,
y = 3,
z=1
As an order triple: (1, 3, 1)
Solve the system.
2x – y – 3z = 1
4x + 3y + 2z = -4
-3x + 2y + 5z = -3
Step 1: Select an equation and solve it for a
variable. Equation 1 would be a good choice
because the y has a coefficient of 1.
2x – y – 3z = 1
y = 2x – 3z – 1
Step 2: Substitute this equation into the other two
equations. Simplify.
4x + 3y + 2z = -4
-3x + 2y + 5z = -3
4x + 3(2x – 3z – 1) + 2z = -4
-3x + 2(2x – 3z – 1) + 5z = -3
4x + 6x – 9z – 3 + 2z = -4
-3x + 4x – 6z – 2 + 5z = -3
10x – 7z = -1
x – z = -1
Now we have a system of two equations with two
variables.
10x – 7z = -1
x – z = -1
Step 3: Solve one of the equations for a variable.
Choose the 2nd and solve for x.
x – z = -1
x=z–1
Step 4: Substitute this equation into the other.
Solve for the variable.
10x – 7z = -1
10(z – 1) – 7z = -1
10z – 10 – 7z = -1
3z = 9
z=3
Step 5: Substitute the value of z into the equation
for x.
x=z–1
x = (3) – 1
x=2
Step 6: Substitute the values of x and z into the
equation for y.
y = 2x – 3z – 1
y = 2(2) – 3(3) – 1
y=4–9–1
y = -6
The solution to the system is:
x = 2,
y = -6,
z=3
As an order triple: (2, -6, 3)
Solve the system.
3x – y + 4z = -10
-x + y + 2z = 6
2x – y + z = -8
Step 1: Select an equation and solve it for a
variable. Any equation would work because they
all have a variable with a coefficient of 1. Let’s
choose the 2nd and solve for y.
-x + y + 2z = 6
y = x – 2z + 6
Step 2: Substitute this equation into the other two
equations. Simplify.
3x – y + 4z = -10
2x – y + z = -8
3x – (x – 2z + 6) + 4z = -10
2x – (x – 2z + 6) + z = -8
3x – x + 2z – 6 + 4z = -10
2x – x + 2z – 6 + z = -8
2x + 6z = -4
x + 3z = -2
Now we have a system of two equations with two
variables.
2x + 6z = -4
x + 3z = -2
Step 3: Solve one of the equations for a variable.
Choose the 2nd and solve for x.
x + 3z = -2
x = -3z – 2
Step 4: Substitute this equation into the other. Solve
for the variable.
2x + 6z = -4
2(-3z – 2) + 6z = -4
-6z – 4 + 6z = -4
-4 = -4
This equation is always true. This means that the value
of z does not impact the solution. It could be literally
any and every number. Therefore, there are infinitely
many solutions.
One solution: The three planes
intersect at exactly one point.
No solution: The three planes
do not all intersect at the
same place.
Infinitely many solutions: The
intersection of the three
planes is an infinitely long line
(or possibly plane).